Java的String类中length（）函数的复杂性是什么？

Question

Is it O(n) or O(1) (By saving length in a private variable during string allocation to the object).

if it is O(n), does it mean that the complexity of following code is O(n^2)?

for(int i=0; i<s.length()-1;i++){
    //some code here!
}

Answer 1

It is O(1) as the length is already known to String instance.

From JDK 1.6 it is visible.

public int length() {
    return count;
}

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Update

It is important to understand why they can cache the value of count and keep using same value for count . The reason lies in a great decision they took when designing String , its Immutability.

Answer 2

In Java any String is backed up by an final array. So it is simple to just return the array length. So it is O(1) complexity. And if you think in your code

for(int i=0; i<s.length()-1;i++){
    //some code here!
}

s.length() is called for every iteration then you are not right. Modern compiler optimizes this type of call and changes s.length() to constant number(ie the length of the String instance).

Answer 3

The complexity is O(1) Since String class have the length as a field . It's not O(n^2).

Answer 4

String内部维护一个字符数组 ， 数组的长度是数组对象的属性 ，因此O（1）作为其简单的属性读取。